Question

A ball is launched from ground level with an initialupwards velocity of 20 m/s and an initial horizontalvelocity of 30 m/s. How far from its startingposition does the ball land assuming the ground is level?

Answer 1

Givens.

• The vertical initial velocity is 20 m/s.

,

• The horizontal initial velocity is 30 ms/s.

,

• The gravity is -9.8 m/s^2.

Use the following formula to find the horizontal reach of the projectile.

`R=(v^2_0\sin 2\theta)/(g)`

First, we need to find the initial speed using the components we already have.

`\begin{gathered} v_0=\sqrt[]{v^2_x+v^2_y}=\sqrt[]{30^2+20^2}=\sqrt[]{900+400}=\sqrt[]{1300} \\ v_0\approx36((m)/(s)) \end{gathered}`

Now we need to find the direction of the initial velocity.

`\begin{gathered} \theta=\tan ^(-1)((y)/(x))=\tan ^(-1)((20)/(30)) \\ \theta\approx33.69 \end{gathered}`

Once we have the initial velocity and its direction, use the formula to find the horizontal reach.

`\begin{gathered} R=\frac{(\sqrt[]{1300})^2\cdot\sin (2\cdot33.7)}{-9.8} \\ R=(1300\cdot(-0.99))/(-9.8) \\ R=(-1286.50)/(-9.8) \\ R=131.28m \end{gathered}`

Therefore, the horizontal reach is 131.28 meters, approximately.